Systems of Three Variables 2 Systems of Three Variables 2
Systems of Three Variables 2
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- Solve this system. And once again, we have three equations
- with three unknowns, so this is essentially trying to find out where
- three different planes would intersect in three dimensions.
- And to do this, if you want to do by elimination, if we want to eliminate variables
- it looks like -- well, it looks like we have a negative z here we have
- a plus 2z we have a 5z over here.
- If we were to scale up this third equation by positive 2,
- then you'd have a negative 2z here and would cancel out
- with this 2z there, and if you were to scale it up by 5, you'd
- have a negative 5z here and that can cancel out with that 5z over there.
- So let's try to cancel out; let's try to eliminate the z's first.
- So let me start with this equation up here;
- I'll just rewrite it, so we have -- I'll draw an arrow over here --
- we have x + 2y + 5z is equal to -17, and then to cancel out or to eliminate the z's,
- I'll multiply this equation here times 5.
- So I'm going to multiply this equation times 5. So we have to multiply both sides by 5.
- So 3x times 5 is 15x, y times 5 is plus 5y, and then -z times 5 is -5z --
- that's the whole point why we're multiplying by 5 -- is equal to 3 times 5
- which is equal to 15. And so if we add these two equations,
- we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z --
- those are going to cancel out -- and that is going to be equal to
- negative 17 plus 15 is negative 2.
- So we're able to use the constraints in that equation and that equation
- and now we have the equation with just x and y.
- So let's try to do the same thing -- let's try to eliminate the z's --
- but now use this equation and this equation.
- So this equation -- let me just rewrite it over here --
- we have 2x minus 3y plus 2z is equal to negative 16 and I just rewrote it.
- And now, so that this 2z gets eliminated, let's multiply this equation times 2.
- So let's multiply it times 2 so we'll have a negative 2z here to eliminate with the positive 2z.
- So 2 times 3x is 6x, 2 times y is plus 2y, and 2 times negative z is negative 2z
- is equal to 2 times 3 is equal to 6. And now we can add these two equations.
- 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get cancel out,
- and then that is equal to negative 16 plus 6 is negative -- is negative 10.
- So now we have two equations with two unknowns; we eliminated the z's.
- And, let's see, if we want to eliminate again,
- we have a negative y over here, we have a positive 7y, we could have eliminated the y's
- if we multiply this times 7, and add the two equations. So let's do that.
- So let's multiply this times 7. 7 times 8 is 56, so it's 56x minus 7y is equal to
- 7 times negative 10 is equal to negative 70, and now we can add these two equations.
- I'm now trying to eliminate the y's. So we have 16x plus 56x, that is 72x.
- So we have 72x, these guys eliminate, equal to negative 72.
- Negative 2 plus negative 70. Divide both sides by 72, and we get x is equal to negative 1.
- And now we just have to substitute back to figure out what y and z are equal to.
- So let's go back to this equation right over here,
- we have 8x minus y is equal to negative 10.
- If x is equal to negative 1, that means 8 times negative 1 or negative 8
- minus y is equal to negative 10. We can add 8 to both sides, and so we have
- negative y is equal to negative 2. Or multiplying both sides by negative 1,
- y is equal to 2. Let me square that off. So x is equal to negative 1, y is equal to 2,
- we now just have to worry about z, and we can go back to any of these up here.
- So let's just use -- I'll just use the last numbers just because they seem lower --
- so if we substitute back into the last equation right over here,
- we have 3 times x, which is 3 times negative 1, plus y, which is 2, minus z is equal to 3.
- So negative 3 plus 2 minus z is equal to 3.
- And this is negative 1 minus z is equal to 3, and then add one to both sides,
- we get -- these cancel out -- negative z is equal to 4, multiplying both sides by negative 1,
- you get z is equal to negative 4.
- So we're done. Let's verify that these solutions or this solution of
- x is negative 1, y is equal to 2, z is equal to negative 4 actually satisfies
- all three of these constraints. So let's substitute into the first one.
- So we have x + 2y + 5z. So that is x is negative 1 plus 2 times y, so plus 4,
- plus 5z, so minus 20 has to be equal to negative 17.
- And this is negative, this right here is positive 3 minus 20 is indeed equal to negative 17.
- So it satisfies the first constraint. So the second one, 2 times x, 2 times negative 1, that's negative 2,
- minus 3 times y, that's minus 6, plus 2 times z, z is equal to negative 4,
- so that's 2 times negative 4 is negative 8. That needs to be equal to negative 16.
- Negative 2 minus negative 8 is -- sorry -- negative 2 minus negative 6 is negative 8,
- subtract another 8, you get negative 16. So it meets the second constraint.
- And then finally, let's look at the last constraint. We have 3 times x,
- so negative 3 plus y, so plus 2, minus z, so minus negative 4 is the same thing as plus 4,
- that needs to be equal to 3. So negative 3 plus 2 is negative 1, plus 4 is indeed equal to 3.
- So we found our point of intersection in three dimensions, at these 3 planes.
- x is negative 1, z is negative 4, y is 2, and we're able to verify it that it does
- indeed meet all of the constraints.
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